Question

If A speaks truth with probability $\frac{4}{5}$ and B with $\frac{3}{4}$, find probability they contradict.

• A. $\frac{3}{20}$
• B. $\frac{1}{5}$
• C. $\frac{7}{20}$
• D. $\frac{4}{5}$

Hint:

Contradictions. Contradiction = exactly one speaks the truth; sum of two exclusive cases.

Answer 1

The Correct Option is C

Contradiction = A true, B false OR A false, B true. \[ P = \left(\frac{4}{5} \cdot \frac{1}{4}\right) + \left(\frac{1}{5} \cdot \frac{3}{4}\right) = \frac{1}{5} + \frac{3}{20} = \frac{7}{20} \]

Answer 2

Step 1: Understand the problem
A speaks the truth with probability \(\frac{4}{5}\) and B speaks the truth with probability \(\frac{3}{4}\). We need to find the probability that they contradict each other.

Step 2: Define events
Let:
- \(T_A\) = A speaks the truth, with \(P(T_A) = \frac{4}{5}\)
- \(F_A\) = A lies, with \(P(F_A) = 1 - \frac{4}{5} = \frac{1}{5}\)
- \(T_B\) = B speaks the truth, with \(P(T_B) = \frac{3}{4}\)
- \(F_B\) = B lies, with \(P(F_B) = 1 - \frac{3}{4} = \frac{1}{4}\)

Step 3: When do they contradict?
They contradict if one speaks the truth and the other lies. So the probability of contradiction is:
\[ P(\text{contradiction}) = P(T_A \cap F_B) + P(F_A \cap T_B) \]

Step 4: Calculate probabilities
\[ P(T_A \cap F_B) = P(T_A) \times P(F_B) = \frac{4}{5} \times \frac{1}{4} = \frac{4}{20} = \frac{1}{5} \]
\[ P(F_A \cap T_B) = P(F_A) \times P(T_B) = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20} \]

Step 5: Add the two probabilities
\[ P(\text{contradiction}) = \frac{1}{5} + \frac{3}{20} = \frac{4}{20} + \frac{3}{20} = \frac{7}{20} \]

Final answer: \(\displaystyle \frac{7}{20}\)